Question

A car is moving with a speed of 72km/h brakes are applied to stop it in 4seconds find a)the retardation produced b)the distance travelled by the car before coming to rest

Answer 1

Explanation:

$- 5 \frac{m}{ {sec}^{2} }$

288km

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Answer 2

To answer this question use the formula:
v=u+at
s={(u+v)/2}t
v2=u2+2as
s=ut+1/2at2  where s-displacement, v-final velocity, u-initial velocity, a-acceleration/retardation, t-time
i)How much time does the car take to stop?

in this case V=72km/hr (convert to m/s=20m/s), u=0m/s, a=m/s2
since v=u+at...at=v-u....t=(v-u)/a
therefore t=(20-0)/5.....=20/5 =4s
therefore it takes 4 seconds to stop

ii)distance covered before stopping: displacement (s)

s= [(u+v)/2]t....=0+20/2 x 4....= 10 x 4= 40ms
hence distance covered is 40 meters

iii)what is stopping distance if speed is doubled?

that means v= 40m/s
hence displacement s,
s= [(u+v)/2]t....= 40/2 x 4
    displacement therefore at double speed =  80 meters